The integral $\int \sin ^ { 4 } x \cos ^ { 3 } x d x$ is

A) $\frac { 5 \sin ^ { 5 } x + 7 \sin ^ { 7 } x } { 35 } + C$
B) $\frac { 5 \sin ^ { 5 } x - 7 \sin ^ { 7 } x } { 35 } + C$
C) $- \frac { 7 \sin ^ { 5 } x - 5 \sin ^ { 7 } x } { 35 } + C$
D) $\frac { 7 \sin ^ { 5 } x + 5 \sin ^ { 7 } x } { 35 } + C$
E) $\frac { 7 \sin ^ { 5 } x - 5 \sin ^ { 7 } x } { 35 } + C$

Question

The integral $\int \sin ^ { 4 } x \cos ^ { 3 } x d x$ is

A) $\frac { 5 \sin ^ { 5 } x + 7 \sin ^ { 7 } x } { 35 } + C$
B) $\frac { 5 \sin ^ { 5 } x - 7 \sin ^ { 7 } x } { 35 } + C$
C) $- \frac { 7 \sin ^ { 5 } x - 5 \sin ^ { 7 } x } { 35 } + C$
D) $\frac { 7 \sin ^ { 5 } x + 5 \sin ^ { 7 } x } { 35 } + C$
E) $\frac { 7 \sin ^ { 5 } x - 5 \sin ^ { 7 } x } { 35 } + C$

Quizplus · Accepted Answer

To solve the integral $\int \sin ^ { 4 } x \cos ^ { 3 } x d x$, use the substitution $u = \sin x$, which leads to $du = \cos x dx$. The integral becomes $\int u^4 (1-u^2) du$, which simplifies to $\int (u^4 - u^6) du$. Integrating term by term gives $\frac{u^5}{5} - \frac{u^7}{7} + C$, or $\frac{7u^5 - 5u^7}{35} + C$, substituting back $u = \sin x$ results in $\frac{7\sin^5 x - 5\sin^7 x}{35} + C$.

The Integral ∫sin⁡4xcos⁡3xdx\int \sin ^ { 4 } x \cos ^ { 3 } x d x∫sin4xcos3xdx

The Integral $\int \sin ^ { 4 } x \cos ^ { 3 } x d x$